Question

lim(x→1)⁡(1/(x-1)+(x-5)/(x²+2x-3))=1/2

Answer 1

Vamos calcular as raízes de $x^2+2x-3$ para fatorá-lo:

$x=\frac{-2\pm\sqrt{2^2-4\cdot(-3)}}{2}$

$x=\frac{-2\pm\sqrt{16}}{2}$

$x=\frac{-2\pm4}{2}$

$x=-1\pm2$

Sendo -3 e 1 as raízes, temos que $x^2+2x-3=(x-1)(x+3)$, logo:

$\frac{1}{x-1}+\frac{x-5}{x^2+2x-3}=\frac{1}{x-1}+\frac{x-5}{(x-1)(x+3)}$

$\frac{1}{x-1}+\frac{x-5}{x^2+2x-3}=\frac{1}{x-1}\cdot\left(1+\frac{x-5}{x+3}\right)$

$\frac{1}{x-1}+\frac{x-5}{x^2+2x-3}=\frac{1}{x-1}\cdot\frac{x+3+x-5}{x+3}$

$\frac{1}{x-1}+\frac{x-5}{x^2+2x-3}=\frac{1}{x-1}\cdot\frac{2x-2}{x+3}$

$\frac{1}{x-1}+\frac{x-5}{x^2+2x-3}=\frac{1}{x-1}\cdot\frac{2(x-1)}{x+3}$

$\frac{1}{x-1}+\frac{x-5}{x^2+2x-3}=\frac{2}{x+3}$

Concluindo assim que:

$\lim_{x\to1}\left(\frac{1}{x-1}+\frac{x-5}{x^2+2x-3}\right)=\lim_{x\to1}\left(\frac{2}{x+3}\right)$

$\lim_{x\to1}\left(\frac{1}{x-1}+\frac{x-5}{x^2+2x-3}\right)=\frac{2}{1+3}=\frac{1}{2}$